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I was reading notion of convergence in random variables and I am not sure if I understood them correctly. According to Wikipedia, for sequence of random variables $X_1,X_2,...$, almost sure convergence means $P(\lim_{n \rightarrow \infty}X_n = X)=1$ which is same as $P(\omega \in \Omega : lim_{n \rightarrow \infty} X_n(\omega) = X(\omega))= 1 $. Now as random variables are function from sample space to $R$, then it seems almost sure convergence means that we have pointwise convergence of functions($X_n$'s to $X$'s) except for may be at sets of measure zero? Now I can see how this implies convergence in distribution but how about random variable $Y=lim_{n \rightarrow \infty}X_n$ and $X$, can we say $Y=X$? I can not see how this is true. Because although we have same functions(random variables), underlying randomness(which could be different for Y and X) could generate different $\omega$'s which could then be mapped by $Y$ and $X$ to different values in $R$. Also if this is not true then how is almost sure convergence different from convergence in distribution? Can we have some non-trivial example where the limiting random variable is not constant. I don't have mathematics background so please try to avoid measure theoretic language and sorry for the long and confusing post. Thanks a lot in advance.

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  • $\begingroup$ If you define $Y = \lim_{n\rightarrow\infty} X_n$ (assuming the limit exists) then indeed $Y=X$ almost surely. It does not make sense to say "underlying randomness is different for $Y$ and $X$" since we have $Y(\omega)$ and $X(\omega)$. They are defined for the same $\omega \in \Omega$. $\endgroup$
    – Michael
    Nov 22 '17 at 23:08
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    $\begingroup$ An example of convergence in distribution is if $\{X_n\}$ are i.i.d Gaussian $N(0,1)$. Then $X_n$ converges in distribution to $X_1$ (it does not converge almost surely). An example of almost sure convergence is if $Z_n= X_1 + 1/n$ for all $n$, then indeed $Z_n\rightarrow X_1$ almost surely (in fact, surely). Another example is $W_n = X_1 + X_n/n$, then indeed $W_n\rightarrow X_1$ almost surely (but not surely). $\endgroup$
    – Michael
    Nov 22 '17 at 23:09
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If $Y$ is another limit with probability $1$ then $X =Y$ with probability $1$, that is $P(X\not = Y)=0$, that is, there is a subset of $\Omega$ whose probability is $0$ and if we omit this set then we can say that $X(\omega)=Y(\omega)$ for all $\omega$ (not in that set).

As far as the relationship between the two concept of convergence. Take a specific example over $\Omega=[0,1]$. Let the probability measure be uniform (fined by the length). Consider $X(\omega)=\omega$. This is going to be the limiting random variable. And define the $X_n$s as functions from $[0,1]$ to $R$ the following "walking" way. First divide $[0,1]$ into four subintervals: $[0,1/4],(1/4,1/2],(1/2,3/4],(3/4,1]$. Let $X_1=0$ over $[0,1/4]$ and $\omega$ otherwise. Let $X_2=0$ over $(1/4,1/2]$ and $\omega$ otherwise and so on. Then halve these intervals and define $X_5=0$ over $[0,1/8]$ and $\omega$ otherwise. Then let the abyss "walk" through the $8$ intervals.

The figure below depicts $X_{10}$ and $X_{11}$ the $10$th and the $11$th elements of this walking abyss.

enter image description here

Then halve the current intervals and let walk the next series the similar way. Then halve again. Now, you are going to have narrower and narrower abysses walking through the interval $[0,1]$. The probability that $X_n=\omega$ is getting larger and larger because the interval over which $X_n$ is $0$ is getting narrower and narrower. So, this is convergence in distribution to $X=\omega$. Do we have convergence with probability $1$? No! Because for any $\omega$ there will be always an $n$ for which the slimmer and slimmer walking abyss will make $X_n=0$ for that $\omega$. What is more, $P(\lim X_n(\omega)=\omega)=0$.

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  • $\begingroup$ This example is way too complicated to just show convergence in distribution holds but not almost sure convergence. In fact this example is showing a more complex situation where convergence "in probability" holds but not almost surely. $\endgroup$
    – Michael
    Nov 22 '17 at 23:29
  • $\begingroup$ @Michael: OK. Right. But I don't agree with the "way too complicated" part. $\endgroup$
    – zoli
    Nov 22 '17 at 23:35
  • $\begingroup$ Compared to an iid sequence of Gaussian variables $N(0,1)$? $\endgroup$
    – Michael
    Nov 22 '17 at 23:36
  • $\begingroup$ @Michael: Yes, right. My reason to add the more complicated and way deeper example is that, in my opinion the OP were having something like this in mind. But, yes. If you take the question word by word then you are very right. $\endgroup$
    – zoli
    Nov 22 '17 at 23:38
  • $\begingroup$ This is certainly an important and deeper example. (+1 for that, and nice picture). $\endgroup$
    – Michael
    Nov 22 '17 at 23:39

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